
Lemma 10.147.5. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p$ in $R$. If $S/R$ is unramified at $\mathfrak q$ then

1. we have $\mathfrak p S_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and

2. the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite separable.

Proof. We may first replace $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ and assume that $R \to S$ is unramified. The base change $S \otimes _ R \kappa (\mathfrak p)$ is unramified over $\kappa (\mathfrak p)$ by Lemma 10.147.3. By Lemma 10.138.3 it is smooth hence étale over $\kappa (\mathfrak p)$. Hence we see that $S \otimes _ R \kappa (\mathfrak p) = (R \setminus \mathfrak p)^{-1} S/\mathfrak pS$ is a product of finite separable field extensions of $\kappa (\mathfrak p)$ by Lemma 10.141.4. This implies the lemma. $\square$

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